A) (4,4)
B) (-1,2)
C) (9/4,3/8)
D) None of these
Correct Answer: C
Solution :
\[2y\frac{dy}{dx}=x\,\{-2(2-x)\}+{{(2-x)}^{2}}\cdot 1\] \[=(2-x)\,(2-x-2x)\] \[=(2-x)\,(2-3x)\] \[{{\left( \frac{dy}{dx} \right)}_{(1,\,1)}}=\frac{(2-1)\,(2-3)}{2}=-\frac{1}{2}\] \[\therefore \] Equation of tangent at \[(1,\,1)\] is \[y-1=-\frac{1}{2}\,(x-1)\] \[\Rightarrow \] \[y=1-\frac{x}{2}+\frac{1}{2}=\left( \frac{3}{2}-\frac{x}{2} \right)\] \[\Rightarrow \] \[{{y}^{2}}=\frac{{{(3-x)}^{2}}}{4}\] \[\Rightarrow \] \[x{{(2-x)}^{2}}=\frac{{{(3-x)}^{2}}}{4}\] \[[\because \,\,{{y}^{2}}=x{{(2-x)}^{2}}]\] After solving, we get \[x=9/4\] \[\therefore \] \[y=\frac{3}{8}\] So, P is \[\left( \frac{9}{4},\,\frac{3}{8} \right)\]
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